What is the solution for y^{''}-y^'-6y=6t^2 ?

The solution for y^{''}-y^'-6y=6t^2 is y=c_{1}e^{3t}+c_{2}e^{-2t}-t^2+t/3-7/18

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y^{''}-y^'-6y=6t^2

y^{^{''}} - y^{^{'}} - 6 y = 6 t^{2}

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - t^{2} + \frac{t}{3} - \frac{7}{18}

Solution steps

y^{''} (t) - y^{'} (t) - 6 y = 6 t^{2}

Solve linear ODE:

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - t^{2} + \frac{t}{3} - \frac{7}{18}

y = c_{1} e^{3 t} + c_{2} e^{- 2 t} - t^{2} + \frac{t}{3} - \frac{7}{18}

y^{^{''}} - 3 y^{^{'}} = 4 sin (x)

y^{^{''}} - y^{^{'}} + y = t^{2} + cos (t)

y^{^{''}} - 2 y^{^{'}} - 3 y = 6 x e^{3 x}

y^{^{''}} + 4 y^{^{'}} + 3 y = e^{5 x}

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y + x^{2} = - 1

What is the solution for y^{''}-y^'-6y=6t^2 ?
The solution for y^{''}-y^'-6y=6t^2 is y=c_{1}e^{3t}+c_{2}e^{-2t}-t^2+t/3-7/18